91Ó°ÊÓ

The given data is about the survey based on the likelihood of the terrorist attacks in their community

The objective is to find the confident interval of the context

If the proportion of observations in a random sample of size $n$that belong to a class of interest is $\hat{p}$, an approximate $100(1-\mathrm{Î\pm })\%$confidence interval on the proportion $p$of the population that belongs to this class is$\hat{p}$.

$\hat{p}-z_{\frac{\mathrm{Î\pm }}{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}≤p≤\hat{p}+z_{\frac{\mathrm{Î\pm }}{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

where $z_{\frac{a}{2}}$is the upper$\frac{\mathrm{Î\pm }}{2}$percentage point of the standard normal distribution.

Nine $x=9$ of $n=48$respondents in a recent Zogby International Poll evaluated the likelihood of a terrorist attack in their neighborhood as "likely" or "extremely likely."

If we utilize the plus-four approach, $n^{\text{'}}=n+4=52$ and $x^{\text{'}}=x+2=11$ are the results.

As a result, the point estimate for the population proportion $p$ is $\hat{p}=\frac{x^{\text{'}}}{n^{\text{'}}}=\frac{11}{52}=0.21.$

For $97\%$ two-sided confidence interval,

and

As a result, a $97\%$two-sided $CL$ estimate for the proportion of American adults who believe a terrorist attack in their town is likely or very likely is derived from Equations.

It is simplified as,

$0.089≤p≤0.334$